Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $q = \dfrac{y + 2}{-7y - 49} \div \dfrac{8y^2 + 48y}{y^2 + 13y + 42} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{y + 2}{-7y - 49} \times \dfrac{y^2 + 13y + 42}{8y^2 + 48y} $ First factor the quadratic. $q = \dfrac{y + 2}{-7y - 49} \times \dfrac{(y + 7)(y + 6)}{8y^2 + 48y} $ Then factor out any other terms. $q = \dfrac{y + 2}{-7(y + 7)} \times \dfrac{(y + 7)(y + 6)}{8y(y + 6)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (y + 2) \times (y + 7)(y + 6) } { -7(y + 7) \times 8y(y + 6) } $ $q = \dfrac{ (y + 2)(y + 7)(y + 6)}{ -56y(y + 7)(y + 6)} $ Notice that $(y + 6)$ and $(y + 7)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ (y + 2)\cancel{(y + 7)}(y + 6)}{ -56y\cancel{(y + 7)}(y + 6)} $ We are dividing by $y + 7$ , so $y + 7 \neq 0$ Therefore, $y \neq -7$ $q = \dfrac{ (y + 2)\cancel{(y + 7)}\cancel{(y + 6)}}{ -56y\cancel{(y + 7)}\cancel{(y + 6)}} $ We are dividing by $y + 6$ , so $y + 6 \neq 0$ Therefore, $y \neq -6$ $q = \dfrac{y + 2}{-56y} $ $q = \dfrac{-(y + 2)}{56y} ; \space y \neq -7 ; \space y \neq -6 $